Problem 6 - College Algebra and Calculus: An Applied Approach

Question

The radius \(r\) of a sphere is increasing at a rate of 3 inches per minute. Find the rates of change of the volume when (a) \(r=6\) inches and (b) \(r=24\) inches.

Step-by-Step Solution

Answer

(a) When \(r=6\) inches, the volume is changing at a rate of \(216\pi\) cubic inches per minute. (b) When \(r=24\) inches, the volume is changing at a rate of \(6912\pi\) cubic inches per minute.

Step 1: Write out the given rate of change

The rate of change of the radius \(r\) is given as \(\frac{dr}{dt} = 3\) inches per minute.

Step 2: Differentiate the volume equation

Differentiate the volume of the sphere with respect to time \(t\) to get \(\frac{dV}{dt} = \frac{d[4/3\pi r^3]}{dt} = 4\pi r^2 \frac{dr}{dt}\).

Step 3: Substitute the knowns for \(r=6\)

Plug \(r=6\) and \(\frac{dr}{dt}=3\) into this equation to solve for \(\frac{dV}{dt}\) for this specific radius. This results in \(\frac{dV}{dt} = 4\pi (6)^2 (3) = 216\pi\) cubic inches per minute.

Step 4: Substitute the knowns for \(r=24\)

Similarly for \(r=24\), substitute into the same equation to get \(\frac{dV}{dt} = 4\pi (24)^2 (3) = 6912\pi\) cubic inches per minute.

Key Concepts

Sphere Volume

The volume of a sphere is an essential concept in geometry, particularly when dealing with real-world applications involving spherical objects. The formula for the volume of a sphere is given by \[ V = \frac{4}{3} \pi r^3 \]where \(V\) represents the volume and \(r\) is the radius of the sphere.
This formula shows how the volume changes with the cube of the radius, meaning even small changes in \(r\) can significantly impact the volume.
Understanding this relationship is crucial, especially when analyzing how spheres expand or contract, such as balloons inflating or celestial bodies growing. Whether you're measuring the volume of basketballs or planetary bodies, knowing how to apply this formula accurately is key.

Calculus Applications

Calculus is a powerful mathematical tool used for a wide range of applications, including understanding how different physical quantities change over time. In the context of this exercise, we're interested in how the rate of change of the radius of a sphere affects its volume.
By using calculus, we can find the rate at which the volume of the sphere is changing, which is done by differentiating the volume with respect to time. This is particularly useful in physics and engineering when predicting changes in systems that vary over time. Calculus provides:

• Tools for solving real-world problems involving motion and change
• Methods to find instantaneous rates of change, known as derivatives
• Insight into relationships between variables and their rates of change

Overall, calculus applications help us make sense of dynamic processes and are indispensable in the fields of science and technology.

Differentiation

Differentiation is a key concept in calculus that focuses on finding the rate at which one variable changes with respect to another. In this exercise, we use differentiation to determine how quickly the volume of the sphere is increasing as the radius grows.
The process involves differentiating the volume formula of the sphere, which involves applying basic differentiation rules to \[ V = \frac{4}{3} \pi r^3 \]This yields the derivative \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]which tells us the rate of change of the volume over time.Understanding how to differentiate:

• Allows us to calculate the derivative of functions, crucial for understanding rates of change
• Helps identify how different changes impact a system dynamically
• Is fundamental for solving problems involving motion and growth

Mastering differentiation provides a foundation for solving complex real-world problems where variables interact and change.